Page:Annales de mathématiques pures et appliquées, 1828-1829, Tome 19.djvu/201

$mX=\operatorname {S} x\mathrm {D} m,\qquad mY=\operatorname {S} y\mathrm {D} m,\qquad mZ=\operatorname {S} z\mathrm {D} m\,;$

d’où on conclura

$\operatorname {S} x'\mathrm {D} m=0,\qquad \operatorname {S} y'\mathrm {D} m=0,\qquad \operatorname {S} z'\mathrm {D} m=0.$

Au moyen de ces valeurs, le premier membre de l’équation (5) deviendra

$\operatorname {S} \left\{{\begin{aligned}&{\frac {\operatorname {d} ^{2}X\delta X+\operatorname {d} ^{2}Y\delta Y+\operatorname {d} ^{2}Z\delta Z+\operatorname {d} ^{2}X\delta x'+\operatorname {d} ^{2}Y\delta y'+\operatorname {d} ^{2}Z\delta z'}{\operatorname {d} t^{2}}}\\\\+&{\frac {\operatorname {d} ^{2}x'\delta X+\operatorname {d} ^{2}y'\delta Y+\operatorname {d} ^{2}z'\delta Z+\operatorname {d} ^{2}x'\delta x'+\operatorname {d} ^{2}y'\delta y'+\operatorname {d} ^{2}z'\delta z'}{\operatorname {d} t^{2}}}\end{aligned}}\right\}\mathrm {D} m.$

Maintenant les valeurs de $X,Y,Z$ étant les mêmes pour toutes les molécules $\mathrm {D} m,$ ces quantités et leurs différentielles pourront se mettre hors du signe $\operatorname {S} ,$ ce qui nous donnera

$\operatorname {S} \left({\frac {\operatorname {d} ^{2}X\delta X+\operatorname {d} ^{2}Y\delta Y+\operatorname {d} ^{2}Z\delta Z}{\operatorname {d} t^{2}}}\right)\mathrm {D} m=m\left({\frac {\operatorname {d} ^{2}X\delta X+\operatorname {d} ^{2}Y\delta Y+\operatorname {d} ^{2}Z\delta Z}{\operatorname {d} t^{2}}}\right)$

${\begin{aligned}&\operatorname {S} {\frac {\operatorname {d} ^{2}X}{\operatorname {d} t^{2}}}\delta x'\mathrm {D} m={\frac {\operatorname {d} ^{2}X}{\operatorname {d} t^{2}}}\operatorname {S} \delta x'\mathrm {D} m,\\\\&\operatorname {S} {\frac {\operatorname {d} ^{2}Y}{\operatorname {d} t^{2}}}\delta y'\mathrm {D} m={\frac {\operatorname {d} ^{2}Y}{\operatorname {d} t^{2}}}\operatorname {S} \delta y'\mathrm {D} m,\\\\&\operatorname {S} {\frac {\operatorname {d} ^{2}Z}{\operatorname {d} t^{2}}}\delta z'\mathrm {D} m={\frac {\operatorname {d} ^{2}Z}{\operatorname {d} t^{2}}}\operatorname {S} \delta z'\mathrm {D} m,\\\\&\operatorname {S} {\frac {\operatorname {d} ^{2}x'}{\operatorname {d} t^{2}}}\delta X\mathrm {D} m=\delta X\operatorname {S} {\frac {\operatorname {d} ^{2}x'}{\operatorname {d} t^{2}}}\mathrm {D} m,\\\\&\operatorname {S} {\frac {\operatorname {d} ^{2}y'}{\operatorname {d} t^{2}}}\delta Y\mathrm {D} m=\delta Y\operatorname {S} {\frac {\operatorname {d} ^{2}y'}{\operatorname {d} t^{2}}}\mathrm {D} m,\end{aligned}}$